# Why are the differentiation/integration rules what they are?

So I understand what rules you use where, and the general forms of the rules like:

$$\left(\frac{d}{dx}\right)^nx^k=\frac{k!}{(k-n)!}x^{k-n}$$

My question is why are these the formulas that give us the answers we want? I learned integrals and derivatives using the limit method as the subdivisions got smaller and smaller, but I don't have a good conceptual understanding of the geometric manipulation taking place that allows the above rule to give us (slopes/areas) of exponential curves.

For example, I learned that when you integrate you're finding an bunch of infinitesimal area rectangles under the curve

$$dA=h*dx$$

where the height was the distance from the x-axis to the curve

$$h=f(x)$$

so we get

$$dA=f(x)dx$$

and then to go from an infinitesimal area to the whole area you integrate to get

$$A=F(x)$$

Now from here, if

$$f(x)=x$$

then

$$F(x)=\int{x}dx=\frac{x^2}{2}+C$$

And my question is why does changing the exponent of the curve in this way (add one, divide by new exponent) give us the geometric area under the curve?

• The case of $\int x\,{\rm d}x$ is a bit simpler than the other exponents: the area of a right triangle is half the area of a square, which is geometrically pretty obvious. – anon Apr 19 '15 at 6:27
• Yes, I grasp that one now, thanks. This is right along the lines of what I'm looking for - a way to visualize the process that leads to these rules giving us geometrically significant values in all cases, or even just in exponential cases. – Alec Rhea Apr 19 '15 at 6:41

The fundamental theorem of calculus states $\frac{d}{dx}\int_a^x f(u)\,{\rm d}u=f(a)$. This makes sense because the limiting quotients $\frac{1}{h} \int_a^{a+h}f(u)\,{\rm d}u$ are the average of $f(u)$ over the interval $[a,a+h]$, which will approach $f(a)$ as $h\to0^+$. One can of course draw rectangles to bound the area (assuming $f>0$ for simplicity) between $a\cdot \min_{a\le u\le a+h}f(u)$ and $a\cdot\max_{a\le u\le a+h}f(u)$.
So the problem of why $\int x^ndx=\frac{x^{n+1}}{n+1}+C$ can be converted from one of integration into one of differentiation; the problem then is why $(x^n)'=nx^{n-1}$ (let's say for $n\ge0$).
The graph of $y=x^n$ is not the only way to think about the function $x^n$. Instead, consider it as the size of a hypercube with side length $x$. Visualize it with $n=3$ since that's easiest. If we extend the side length to $x+h$, we've added a few "panels" to the positive-axis-facing faces of the cube that have thickness $h$ and area $x^2$. In addition there are little edges that have the dimensions $h\times h\times x$ and corners of dimensions $h\times h\times h$. As we calculate the difference in volume between the cube of side length $x$ and the cube of side length $x+h$, the edges and corners will fade away because they have the wrong dimensions, but the face panels, of which there are three (one for each positive axis direction) have the right dimensions and they tell us the derivative will be $3x^2$. The same story happens for higher dimensions $n$.
As for $n<0$, perhaps some more finagling is in order. One can visually justify the product rule for positive-valued functions (again for simplicity) using the same kind of argument with rectangles and dimensionality (mirroring the algebra expanding $(u+\Delta u)(v+\Delta v)$ into the relevant terms and making $\Delta x\to0$) and then apply it to $1=x^n\cdot x^{-n}$ to indirectly compute the derivative $(x^{-n})'$ at least with some geometric inspiration.